Case Study: Auditing a Geostationary Delivery with Vis-Viva

"The launch is only half the trip. The other half is a satellite, alone, climbing the last thirty thousand kilometers on its own thruster."

Executive Summary

Every large communications satellite flying today reached its geostationary slot the same way: a launch vehicle dropped it not at GEO but onto a stretched-out geostationary transfer orbit (GTO) — an ellipse whose low point skims a few hundred kilometers above the Earth and whose high point just reaches geostationary altitude — and then the satellite's own engine fired at the high point to circularize. This two-part choreography is the single most common maneuver in commercial spaceflight, flown hundreds of times, and it is nothing but the transfer-orbit example of §6.3 made real.

In this case study we audit that delivery from the outside, using only the vis-viva equation and the energy relations of this chapter. We will reconstruct the speed of the satellite at every stage, compute the size of the apogee-kick burn its onboard propulsion must supply, and track the specific orbital energy from parking orbit to GTO to GEO — confirming, with a real mission profile, the delta-v map figures we took on faith in Chapter 3. The numbers below are representative of a Falcon-9-class GTO delivery (Tier 2/3 — rounded and illustrative, since exact values vary by satellite and launch).

Skills applied

• Computing circular velocity and specific energy at LEO, GTO, and GEO (§6.1–6.2).
• Using vis-viva to find the speed at the two ends of a transfer ellipse (§6.3–6.4).
• Sizing a real orbital burn as a difference of vis-viva speeds (§6.4).
• Tracking the energy ledger across a multi-stage delivery, with a sanity check (§6.2).

Background

The mission profile

A geostationary communications satellite must end up in a circular orbit at GEO radius $r_{\text{GEO}} = 42{,}164\ \text{km}$, where it circles once per sidereal day and so appears fixed over a point on the equator. Getting there has three acts:

1. Ascent to a low parking orbit. The launch vehicle reaches a low circular orbit — take $300\ \text{km}$ altitude, $r_1 = 6{,}671\ \text{km}$ — or passes through this state on its way to GTO.
2. Injection onto GTO. A burn near perigee stretches the orbit outward so its apogee reaches GEO altitude. The result is an ellipse with $r_p = 6{,}671\ \text{km}$ and $r_a = 42{,}164\ \text{km}$. On a Falcon-9-class mission the rocket's upper stage supplies this, dropping the satellite onto GTO.
3. The apogee kick to GEO. At the high point of GTO the satellite is moving slowly; it fires its own engine to speed up and round the ellipse into a circle at GEO. This is the satellite's job, paid from the propellant it carries for life.

Our audit computes the speeds and energies at each act. We use $\mu_\oplus = 3.986\times10^{5}\ \text{km}^3/\text{s}^2$ throughout.

Phase 1: The parking orbit — a speed baseline

Start with the low circular parking orbit at $r_1 = 6{,}671\ \text{km}$. Its circular speed is

$$ v_{\text{circ},1} = \sqrt{\frac{\mu_\oplus}{r_1}} = \sqrt{\frac{3.986\times10^{5}}{6{,}671}} = \sqrt{59.75} = 7.730\ \text{km/s}. $$

Its specific orbital energy is $\varepsilon_1 = -\mu/(2r_1) = -398{,}600/(2\times6{,}671) = -29.9\ \text{MJ/kg}$. Hold these two numbers; every later speed is a departure from this baseline.

Phase 2: The transfer orbit (GTO)

The GTO ellipse runs from $r_p = 6{,}671\ \text{km}$ up to $r_a = 42{,}164\ \text{km}$. Its semi-major axis is half the long diameter:

$$ a_{\text{GTO}} = \frac{r_p + r_a}{2} = \frac{6{,}671 + 42{,}164}{2} = 24{,}418\ \text{km}. $$

Its specific energy is already higher (less negative) than the parking orbit's:

$$ \varepsilon_{\text{GTO}} = -\frac{\mu_\oplus}{2\,a_{\text{GTO}}} = -\frac{398{,}600}{2\times24{,}418} = -8.16\ \text{MJ/kg}. $$

Now the two speeds that matter, from vis-viva $v^2 = \mu(2/r - 1/a)$. At perigee ($r = r_p = 6{,}671\ \text{km}$):

$$ v_p^2 = 3.986\times10^{5}\!\left(\frac{2}{6{,}671} - \frac{1}{24{,}418}\right) = 3.986\times10^{5}\,(2.998\times10^{-4} - 4.095\times10^{-5}) = 103.2\ \text{km}^2/\text{s}^2, $$

so $v_p = 10.16\ \text{km/s}$. At apogee ($r = r_a = 42{,}164\ \text{km}$):

$$ v_a^2 = 3.986\times10^{5}\!\left(\frac{2}{42{,}164} - \frac{1}{24{,}418}\right) = 3.986\times10^{5}\,(4.743\times10^{-5} - 4.095\times10^{-5}) = 2.58\ \text{km}^2/\text{s}^2, $$

so $v_a = 1.61\ \text{km/s}$. The satellite races through perigee at over $10\ \text{km/s}$ and dawdles through apogee at barely $1.6\ \text{km/s}$ — the same six-fold spread we found in the chapter, and the reason the apogee is the smart place to circularize (a small nudge there changes the orbit a lot).

The injection burn (for reference). Going from the parking orbit ($v_{\text{circ},1} = 7.730\ \text{km/s}$) onto GTO requires speeding up at perigee to $v_p = 10.16\ \text{km/s}$, a burn of $\Delta v_{\text{inj}} = 10.16 - 7.73 = 2.43\ \text{km/s}$. This matches the delta-v map's "LEO → GTO $\approx 2.5\ \text{km/s}$" from Chapter 3. On our reference mission the launch vehicle supplies this; we compute it to confirm the map.

Phase 3: The apogee kick to GEO

At GTO apogee the satellite is at GEO altitude but moving far too slowly ($1.61\ \text{km/s}$) to hold a circular orbit there. A circular GEO requires

$$ v_{\text{circ,GEO}} = \sqrt{\frac{\mu_\oplus}{r_{\text{GEO}}}} = \sqrt{\frac{3.986\times10^{5}}{42{,}164}} = \sqrt{9.454} = 3.075\ \text{km/s}. $$

So the satellite's own engine must supply the apogee-kick burn

$$ \Delta v_{\text{AK}} = v_{\text{circ,GEO}} - v_a = 3.075 - 1.61 = 1.47\ \text{km/s}. $$

That is the number the satellite's propellant load is sized against — and it matches the delta-v map's "GTO → GEO $\approx 1.5\ \text{km/s}$." The satellite carries exactly enough propellant to buy this $1.47\ \text{km/s}$ (plus a decade of station-keeping), sized through the rocket equation of Chapter 3.

🔧 Engineering Reality: the plane change hides here. A GTO launched from Cape Canaveral is tilted about $27^\circ$ out of the equatorial plane, but GEO is exactly equatorial. So the real apogee kick must also swing the orbit into the equatorial plane — a plane change combined with the speed-up. Done together at apogee (where the satellite moves slowly, so turning is cheap), the combined burn is about $1.8\ \text{km/s}$, not $1.47$. We deliberately leave the plane-change math to Chapter 10; the point here is that vis-viva gives the energy part of the burn exactly, and doing the turn at slow-moving apogee is another face of "high is slow, and slow is a good place to maneuver."

Phase 4: The energy ledger and a sanity check

Let us track the specific orbital energy through the whole delivery and check that it only ever increases when the engine fires (energy in) and holds constant while coasting (energy conserved):

Read down the energy column: $-29.9 \to -8.16 \to -4.73\ \text{MJ/kg}$. Each burn raises the energy (the orbit climbs the well); the coast across GTO changes the speed enormously ($10.16$ down to $1.61\ \text{km/s}$) but leaves the energy untouched at $-8.16\ \text{MJ/kg}$, exactly as conservation demands. The total energy added by the two burns is $-4.73 - (-29.9) = 25.2\ \text{MJ/kg}$ — the price, in energy, of lifting a kilogram from a low parking orbit to a geostationary slot.

Sanity check. The final GEO speed came out to $3.08\ \text{km/s}$ two independent ways: as $v_{\text{circ,GEO}} = \sqrt{\mu/r_{\text{GEO}}}$ (Phase 3), and as the apogee speed $1.61\ \text{km/s}$ plus the kick $1.47\ \text{km/s}$. They agree, which they must — the burn was designed to reach circular speed. When two routes to the same number agree, the arithmetic is almost certainly right. When they don't, you have found a mistake before it flew.

Discussion Questions

1. The satellite coasts from $10.16\ \text{km/s}$ at perigee to $1.61\ \text{km/s}$ at apogee without firing its engine. Where did the missing $8.5\ \text{km/s}$ of speed "go"? Answer in the language of the two energies.
2. Why is the apogee — not the perigee — the right place to perform the circularization burn? (Think about how much the orbit changes per m/s of burn at each end.)
3. The perigee injection ($2.43\ \text{km/s}$) is larger than the apogee kick ($1.47\ \text{km/s}$), yet both are "half" of the same transfer. Why are they unequal?
4. Using the table, explain how a satellite can have more total energy at GEO than in GTO yet move slower at GEO than at GTO perigee.

Your Turn: Extensions

• Option A (analysis). Redo the audit for a "supersynchronous" transfer whose apogee overshoots to $r_a = 90{,}000\ \text{km}$ before dropping back to circularize at GEO. Compute the new GTO apogee speed and the apogee burn. Is the total delta-v higher or lower than the standard case? (This previews why some operators overshoot to make the plane change cheaper.)
• Option B (computation). Write a Python function transfer(mu, r1, r2) that returns the two burns of a Hohmann-style transfer between circular orbits $r_1$ and $r_2$ using vis-viva, and reproduce the $2.43$ and $1.47\ \text{km/s}$ figures. Do not run it — hand-trace and add # Expected output:.
• Option C (your mission). If your mission (Track A/B/C/D) involves a transfer between two circular orbits, run this same three-phase audit for it and record every speed and the total delta-v in your MDR.

Key Takeaways

1. A GEO delivery is a transfer ellipse plus a circularization burn — exactly the §6.3 example, flown for real hundreds of times.
2. Vis-viva sizes every burn as a difference of speeds: the injection is $v_p - v_{\text{circ},1}$, the apogee kick is $v_{\text{circ,GEO}} - v_a$. No other equation is needed for the energy part.
3. Energy rises only when the engine fires; coasting conserves it. Tracking $\varepsilon$ across the delivery is a powerful check — and it reveals that GEO has more energy but less speed than GTO perigee.
4. The map was right. Our audit reproduced Chapter 3's "$2.5$" and "$1.5\ \text{km/s}$" legs from first principles — the delta-v map is just vis-viva, tabulated.